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Consider an operator (not necessarily linear) $\mathcal{O}: C^{\infty} (\mathbb{R}) \to C^\infty(\mathbb{R})$ which satisfies the following property for any function $f \in C^\infty(\mathbb{R})$ and any constant function $c \in C^\infty(\mathbb{R})$ \begin{equation} \mathcal{O} \left(f + c \right) = \mathcal{O} \left(f \right) \end{equation}

Must the operator $\mathcal{O}$ be a differential operator? If not, what additional conditions should $\mathcal{O}$ have for it to have to be a differential operator? For example, would $\mathcal{O}$ being a linear operator suffice?

Ishan Deo
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    Without any additional property, there is no reason for values on different "lines" ${f + c\mid c \text{ constant}}$ to interfere with each other, thus you should not expect any structure as is. – Bruno B Feb 14 '24 at 17:53

1 Answers1

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No; there are many many linear operators which satisfy this condition. For example, let $\mathcal{O}$ be the operator which sends a smooth function $f$ to the constant function with value $f(1)-f(0)$.

See here for a characterization of the derivative, which then leads to a characterization of differential operators.

diracdeltafunk
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